1) $xy+2x-y=5$
$⇔x.(y+2)-(y+2)=3$
$⇔(y+2)(x-1)=3$
TH1 : $y+2=3,x-1=1 ⇒y=1,x=2$
TH2 : $y+2=-3,x-1=-1⇒y=-5,x=0$
TH3 : $y+2=1,x-1=3⇒y=-1,x=4$
TH4 : $y+2=-1,x-1=-3⇒y=-3,x=-2$
2) Ta có : $2x=3y ⇒ x/3 = y/2 ⇒ x/15=y/10$
$4y=5z⇒y/5=z/4⇒y=10=z/8$
$⇒ \frac{x}{15} = \frac{y}{10} = \frac{z}{8} $
$⇒\frac{4x}{60} = \frac{3y}{30} = \frac{5z}{40} = \frac{7}{70} = \frac{1}{10}$
( Tính chất dãy tỉ số bằng nhau )
$⇒x=1 , y = \frac{3}{2},z=\frac{15}{8} $